2.1.34 \(u_t+u u_x=x\) with \(u(x,0)=f(x)\) Example 3.5.11 in Lokenath Debnath.

problem number 34

Added June 2, 2019.

From example 3.5.11, page 219 nonlinear pde’s by Lokenath Debnath, 3rd edition.

Solve for \(u(x,y)\) \[ u_t+u u_x=x \] with \(u(x,0)=f(x)\)

Mathematica


Failed

Maple


\[u \left (x , t\right ) = \RootOf \left (\left (\mathit {\_Z} \,{\mathrm e}^{2 t}-2 x \,{\mathrm e}^{t}+{\mathrm e}^{2 t} f \left (\mathit {\_Z} \right )+\mathit {\_Z} -f \left (\mathit {\_Z} \right )\right ) \left (\mathit {\_Z} +f \left (\mathit {\_Z} \right )\right )\right ) {\mathrm e}^{t}+{\mathrm e}^{t} f \left (\RootOf \left (\left (\mathit {\_Z} \,{\mathrm e}^{2 t}-2 x \,{\mathrm e}^{t}+{\mathrm e}^{2 t} f \left (\mathit {\_Z} \right )+\mathit {\_Z} -f \left (\mathit {\_Z} \right )\right ) \left (\mathit {\_Z} +f \left (\mathit {\_Z} \right )\right )\right )\right )-x\]

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